Question

A particle is thrown up vertically with a velocity of 50ms -1 . How high would the particle rise and what time would it take to reach the highest point ? (g=10ms -2 )

Answer 1

Given:
Initial velocity = u= 50m/s
g=10m/s²
1) maximum height = h= u²/2g
                                       =50x 50/2x10
                                       =125m
2) time taken to reach maximum height(time of ascent)=t1=u/g⇒50/10
                                                                                             =5sec

Answer 2

✬ Maximum Height = 125 m ✬

✬ Time Taken = 5 Seconds ✬

Explanation:

Given:

• Initial Velocity u = 50m/s
• Final Velocity v = 0 ( at Maximum height )
• G = –10 m/s² ( going upward )

To Find:

• How hight would the particle rise and what time would take to reach highest point?

Solution: We know that :-

★ 2gS = v² – u² ★

• Putting all the values in given formula •

$\small\implies{\sf }$ 2 x –10 x S = 0² – 50²

$\small\implies{\sf }$ –20S = – 2500

$\small\implies{\sf }$ S = –2500/–20

$\small\implies{\sf }$ S = 125 m

Hence, The Maximum height reached by the stone will be 125 m.

→ To find time we will use the formula

★ V = u + gt ★

$\small\implies{\sf }$ 0 = 50 + (–10) x t

$\small\implies{\sf }$ 0 = 50 –10t

$\small\implies{\sf }$ –50 = –10t

$\small\implies{\sf }$ –50/–10 = t

$\small\implies{\sf }$ 5 = t

Hence, The time taken by the particle to reach the maximum height will be 5 Seconds.